Puzzle for September 26, 2021 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
* DE and EF are 2-digit numbers (not D×E or E×F).
Scratchpad
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Hint #1
eq.2 may be written as: A – C + D = E + B + F In the above equation, replace B + F with A + C (from eq.1): A – C + D = E + A + C Subtract A from both sides, and add C to both sides: A – C + D – A + C = E + A + C – A + C which becomes eq.2a) D = E + 2×C
Hint #2
eq.4 may be written as: 10×D + E – (10×E + F) = C + F which is equivalent to 10×D + E – 10×E – F = C + F which becomes 10×D – 9×E – F = C + F Add F to both sides of the above equation: 10×D – 9×E – F + F = C + F + F which becomes eq.4a) 10×D – 9×E = C + 2×F
Hint #3
In eq.4a, substitute (E + 2×C) for D (from eq.2a): 10×(E + 2×C) – 9×E = C + 2×F which becomes 10×E + 20×C – 9×E = C + 2×F which becomes E + 20×C = C + 2×F Subtract C from both sides of the equation above: E + 20×C – C = C + 2×F – C which becomes eq.4b) E + 19×C = 2×F
Hint #4
Since F must be a one-digit non-negative integer, then: F ≤ 9 which means 2×F ≤ 18 Combining the above inequality with eq.4b yields: ie.4c) E + 19×C ≤ 18
Hint #5
To make ie.4c true, check several possible values for C: If C = 0, then E + 19×C = E + 19×0 = E + 0 = E which means E + 19×C ≤ 18 If C ≥ 1, then E + 19×C ≥ E + 19×1 = E + 19 ≥ 18 which means E + 19×C is not ≤ 18 Since E must be a one-digit non-negative integer, the above equations make: C = 0
Hint #6
Substitute 0 for C in eq.4b: E + 19×0 = 2×F which makes E = 2×F
Hint #7
Substitute (2×F) for E, and 0 for C in eq.4a: 10×D – 9×(2×F) = 2×F + 0 which becomes 10×D – 18×F = 2×F Add 18×F to both sides of the above equation: 10×D – 18×F + 18×F = 2×F + 18×F which becomes 10×D = 20×F Divide both sides by 10: 10×D ÷ 10 = 20×F ÷ 10 which makes D = 2×F
Hint #8
In eq.3, substitute 0 for C, and 2×F for E and D (since D ≠ 0 (from eq.3), and 2×F = D, therefore 2×F ≠ 0): 2×F + F + 0 = (B ÷ 2×F) – 0 which becomes 3×F = B ÷ 2×F Multiply both sides of the above equation by 2×F: (3×F) × 2×F = (B ÷ 2×F) × 2×F which becomes eq.3a) 6×F² = B
Hint #9
To make eq.3a true, check several possible values for F and B: If F = 0, then B = 6×F² = 6×0² = 6×0 = 0 If F = 1, then B = 6×F² = 6×1² = 6×1 = 6 If F = 2, then B = 6×F² = 6×2² = 6×4 = 24 If F > 2, then B > 24 Since B must be a one-digit integer, the only two values for F that make eq.3a true are: F = 0 or F = 1 However, F ≠ 0 (since 2×F ≠ 0), therefore: F = 1 making B = 6×F² = 6×1² = 6×1 = 6 (from eq.3a) D = E = 2×F = 2×1 = 2
Solution
Substitute 0 for C, 2 for D and E, 6 for B, and 1 for F in eq.2: A – 0 + 2 = 6 + 2 + 1 which becomes A + 2 = 9 Subtract 2 from each side of the above equation: A + 2 – 2 = 9 – 2 which makes A = 7 and ABCDEF = 760221